# How do you identify all asymptotes or holes for f(x)=(x^2-9)/(2x^2+1)?

Sep 17, 2016

horizontal asymptote at $y = \frac{1}{2}$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: $2 {x}^{2} + 1 = 0 \Rightarrow {x}^{2} = - \frac{1}{2}$

there are no real solutions for x hence there are no vertical asymptotes.

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest power of x that is ${x}^{2}$

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} - \frac{9}{x} ^ 2}{\frac{2 {x}^{2}}{x} ^ 2 + \frac{1}{x} ^ 2} = \frac{1 - \frac{9}{x} ^ 2}{2 + \frac{1}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{1 - 0}{2 + 0}$

$\Rightarrow y = \frac{1}{2} \text{ is the asymptote}$

Holes occur when there are duplicate factors on the numerator/denominator. This is not the case here hence there are no holes.
graph{(x^2-9)/(2x^2+1) [-18.01, 18.04, -9.02, 9]}